Hey, I started to learn what groups are... I asked myself, is the multiplication in R a group?

My counterargument is, that there is no identity element, because 1*e = e*1 = e, expect for e=0. And the second one is, that there is no symmetric element, because: a*b = b * a = e, where b = 1/a and e = the no identity element. This works, except for a=0...

Am I wrong in my argumentation, or is the multiplication in the real set no group?

 

Thank you!

 

The set of all real numbers is a monoid under standard multiplication.

Real numbers without 0 form a group under multiplication.

Shameless plug here, but I use the monoid structure of the real and comment on the use of the group of non-zero reals in my latest preprint (http://arxiv.org/abs/1502.06092).

 

What I use is the fundamental result that any smooth action of the monoid of multiplicative reals on a manifold leads to a particularly nice N-grading of the (structure sheaf of the) manifold. Such actions are called homogeneity structures. In particular we have only functions that have a positive grading and the homogeneous functions form a bases of all functions. We have a kind of polynomial bundle known as a graded bundle. These are the commutative versions of non-negatively graded supermanifolds that you find in the literature, an example would be the so called N-manifolds.

 

Anyway, if you now replace the homogeneity structure with the action of group is non-zero reals, the theory is similar but fundamentally geometrically different. You end up with a principal bundle whose structure group is the group of non-zero reals. Importantly, to every such structure corresponds a line bundle over the base of this principal bundle. In reverse, you can always construct a principal bundle with structure group of the non-zero reals by just forgetting the zero section. There is a one-to-one correspondence.

 

So instead of dealing with line bundles a such, you can consider certain principal bundles and these can be described by a smooth action of the group of non-zero reals. This is something that has not yet be greatly exploited.

 

I thought it maybe nice for Backes to know that these structures are used in modern mathematics.

Edited February 27, 201510 yr by ajb

And just to answer the actual question: the set of real numbers is not a group under the usual multiplication law, but not because there is no neutral element, because 1 is a neutral element under multiplication, but not everyone is invertible, because 0 isn't.

If you put e=0 in your original argument Backes, it still works: 1*0=0*1=0, so 1 is neutral even for 0.

The real numbers are on the other hand a group under the addition law.

Well pitched answer keen +1